Investigation of subalgebra lattices by means of Hasse constants

0.25 MB·English

by  Petr Vojtěchovský

Investigation of Subalgebra Lattices by Means of Hasse
Constants
7
0
0 Petr Vojtˇechovsky´
2

n
a Abstract. Hasseconstantsandtheirbasicpropertiesareintroducedtofacilitatethecon-
J nection between the lattice of subalgebras of an algebra C and the natural action of the
automorphism group Aut(C) on C. These constants are then used to describe the lattice
4
ofsubloopsofthesmallestnonassociativesimpleMoufangloop.
2

]
R
G
1. Introduction
.
h
at To completely describe the lattice of subalgebras Sub(C) of a finite algebra C is
m a difficult task. Moreover, it is not obvious how to store the information about
Sub(C) efficiently, as the cardinality and complexity of Sub(C) is typically much
[
largerthanthatofC. Fortunately,sometimesthereisaprocedurethatallowsusto
1
calculate the join A∨B and meet A∧B for every A, B ∈Sub(C). For instance, it
v
iseasytofindthejoinandmeetinanybooleanalgebraC,although|Sub(C)|grows
4
0 exponentiallyin|C|. It isthis ability to calculate∧and∨thatis oftenunderstood
7 as a complete description of Sub(C).
1 Most of the time we are not so lucky, though, and there is no apparent way to
0
find joins and meets. The main reason is that A∨B and A∧B can be far from
7
both A and B in the lattice Sub(C). It is therefore more convenientto have access
0
/ to a procedure that gives a complete local description of Sub(C). Assuming that
h
it is possible to find all maximal subalgebras of A≤C and all subalgebras B ≤C
t
a in which A is maximal, the lattice Sub(C) can be built up inductively. We will
m
refer to all subalgebras immediately above and immediately below A in Sub(C) as
: neighbors of A, and we denote the set they form by Nbd(A).
v
In this context,it is worthpayingattention to the automorphismgroupAut(C)
i
X anditsnaturalactiononC,sincetheneighborhoodsofAandB willbe“thesame”
r for A, B ∈ Sub(C) belonging to the same orbit of transitivity of Aut(C). Thus,
a
the lattice Sub(C) can be fully described as long as we find
(ℓ ) one representative A from each orbit of Aut(C),
1
(ℓ ) the neighborhood Nbd(A) for every representative A, and
2
1991MathematicsSubjectClassification: Primary: 06B99,20N05. Secondary: 17A75,11H56.
Key words and phrases: Moufangloops,Paigeloops,subalgebralattices,Hasseconstants.
WhileworkingonthispapertheauthorhasbeenpartiallysupportedbyGrantAgencyofCharles
University,grantnumber269/2001/B-MAT/MFF.
1
2 PETRVOJTEˇCHOVSKY´
(ℓ ) an automorphism of C mapping B onto A, for every representative A and
3
every B from the orbit of A.
To save space, we can store subalgebras by their generating sets, and substitute
Nbd(A) and the automorphisms required by (ℓ ) with an efficient algorithm pro-
3
ducing those.
The purpose of this paper is twofold. First, to introduce a general tool—Hasse
constants—that is of some help in all three tasks (ℓ ), (ℓ ), (ℓ ). Secondly, to use
1 2 3
Hasseconstantstodescribethesublooplatticeofthesmallestnonassociativesimple
MoufangloopM∗(2). TheinvestigationofSub(M∗(2))occupiesmostofthispaper,
and is inevitably of rather detailed nature. We maintain that the power of Hasse
constants is sufficiently demonstrated by the fact that Sub(M∗(2)) was not known
before (see Acknowledgement), especially given the importance of M∗(2) for the
real octonions.
Although considerable invention will be required in each particular case, we be-
lievethatHasseconstantswillhelptokeeptrackininvestigationofanysubalgebra
lattice.
A word about the notation: we write C for the cyclic group of order n, D for
n n
thedihedralgroupoforder2n,andE fortheelementaryabelian2-groupoforder
2n
2n. A subalgebra generated by the set S well be denoted by hSi.
2. Hasse constants
Let A, B, C be finite (universal) algebras, A ≤ C. For X ≤ C, let O denote
X
the orbit of X under the natural action of Aut(C) on the set of subalgebras of C
isomorphic to X. We will speak of the subalgebrasof C isomorphic to X as copies
of X in C. Define
HC(B) = |{B0 ≤C; B0 ∼=B}|,
HC(A|B) = |{B0 ≤C; A≤B0 ∼=B}|.
Furthermore, when B ≤C, let
H∗(A|B)=|{B ≤C; A≤B , B ∈O }|.
C 0 0 0 B
In words, H (B) counts the number of copies of B in C, H (A|B) counts the
C C
number of copies of B in C containing A, and H∗(A|B) counts the number of
C
copies of B in C containing A and in the same orbit as B.
Yet another description of these constants is perhaps the most appealing. For
B ≤ C, the constant H (B) counts the number of edges connecting C to a copy
C
of B in the complete Hasse diagram of Sub(C). The remaining two constants can
be interpreted in a similar way. We will therefore refer to them jointly as Hasse
constants.
Note that H (A|B) = H∗(A|B) if Aut(C) acts transitively on the copies of B
C C
in C.
Lemma 2.1. Let A, B, C be algebras, A≤C.
(i) If B′ ∼=B, C′ ∼=C, then HC(B)=HC′(B′).
SUBALGEBRAS AND HASSE CONSTANTS 3
(ii) If A′ ∈OA, B′ ∼=B, then HC(A|B)=HC(A′|B′).
(iii) If A′ ∈O , B ≤C, B′ ∈O , then H∗(A|B)=H∗(A′|B′).
A B C C
Proof. Part (i) is obvious from the definition of H (B). The equality H (A|B)=
C C
HC(A|B′) holds if B ∼= B′. Let A′ ∈ OA, and let f ∈ Aut(C) be an automor-
phism mapping A to A′. Then H (A|B) = H (f(A)|f(B)) = H (A′|f(B)) =
C f(C) C
HC(A′|B). This proves (ii). Part (iii) is similar (use OB =OB′). (cid:3)
Example 2.2. This example shows that the constants H (A|B), H (A′|B) may
C C
differ even though A ∼= A′. Let C be the group C ×C , C = {0, 1}, C = {0,
2 4 2 4
1, 2, 3}, and denote by D = {0, 2} the two-element subgroup of C . The lattice
4
of subgroups of C is depicted in Figure 1. With A =C ×D ∼=C ×C = A′, we
1 2 1
have H (A|C )=26=0=H (A′|C ).
C 4 C 4
C ×C
2 4
C ×D C ×C
2 1 4
h(1, 1)i
C ×C
2 1
h(1, 2)i C ×D
1
C
1
Figure 1. Lattice of subgroups of C ×C
2 4
Proposition 2.3. Let C be an algebra, A, B ≤C. Let A , ..., A be representa-
1 m
tives from all orbits O , ..., O of the action of Aut(C) on the copies of A in
A1 Am
C. Similarly, let B , ..., B be representatives for B. Then
1 n
n
H (A|B) = H∗(A|B ), (1)
C C j
Xj=1
m
H (A)·|O | = |O |·H∗(A |B), (2)
B B Ai C i
Xi=1
m
H (A)·H (B) = |O |·H (A |B). (3)
B C Ai C i
Xi=1
When Aut(C) acts transitively on the copies of B (i.e., when n = 1), then (2)
coincides with (3). When Aut(C) acts transitively on the copies of A (i.e., when
4 PETRVOJTEˇCHOVSKY´
m=1), then
H (A)·|O |=H (A)·H∗(A|B), (4)
B B C C
H (A)·H (B)=H (A)·H (A|B). (5)
B C C C
Proof. Since every copy of B in C belongs to exactly one orbit O , (1) follows.
Bj
To establish (2), count twice the cardinality t of
{(A0, B0); A∼=A0 ≤B0 ∈OB}.
On the one hand,
2.1(i)
t= H (A) = H (A)=H (A)·|O |.
B0 B B B
B0X∈OB B0X∈OB
On the other hand,
m m
t= H∗(A |B)= H∗(A |B)2.1=(iii) |O |·H∗(A |B).
C 0 C 0 Ai C i
A0≤CX,A0∼=A Xi=1A0X∈OAi Xi=1
The proof of (3) is similar to (2). Just count twice the cardinality of the set
{(A , B ); A∼=A ≤B ≤C, B ∼=B}.
0 0 0 0 0
When m=1, (4) and (5) follow immediately from (2) and (3), respectively. (cid:3)
3. Finite simple Moufang loops and loops of type M (G, 2)
2n
Loops satisfying one of the equivalent Moufang identities, for instance the identity
((xy)x)z =x(y(xz)), (6)
arecalledMoufang loops [12]. ByaresultofKunen[9],everyquasigroupsatisfying
(6)isaMoufangloop. Obviously,everygroupisaMoufangloop. Moufangloopsare
power associative (i.e., every 1-generated subloop is a group), in fact diassociative
(i.e., every 2-generated subloop is a group). Every element x of a Moufang loop
has a (unique) two sided inverse x−1.
Paige [11], Doro [6] and Liebeck [10] showed that there is only one class of
nonassociativefinite simple Moufangloops,consistingofloopsM∗(q), one foreach
finite field GF(q).
These loops are best studied via composition algebras. Following [13], let O(q)
be the unique split octonion algebraoverGF(q). Then M∗(q) is isomorphicto the
multiplicative loop of elements of norm 1 in O(q) modulo the center. The algebra
O(q) was first constructed by Zorn as follows. Given a prime power q, let · be the
standard dot product and × the standard vector product on GF(q)3. Then the
algebra of vector matrices
a α
x= (a, b∈GF(q), α, β ∈GF(q)3)
(cid:18) β b (cid:19)
SUBALGEBRAS AND HASSE CONSTANTS 5
with addition defined entry-wise and multiplication governedby
a α c γ ac+α·δ aγ+αd−β×δ
= (7)
(cid:18) β b (cid:19)(cid:18) δ d (cid:19) (cid:18) βc+bδ+α×γ β·γ+bd (cid:19)
is isomorphic to O(q). The norm on O(q) coincides with the determinant detx =
ab−α·β. The neutral element is
1 (0, 0, 0)
e= ,
(cid:18) (0, 0, 0) 1 (cid:19)
and every element x with nonzero norm has inverse
1 b −α
x−1 = · .
detx (cid:18) −β a (cid:19)
The orderof M∗(q) is q3(q4−1)when q is even, andq3(q4−1)/2when q is odd
[11].
Notably, the loop M∗(2) has also connections to the standard real (division)
octonionalgebra. Namely, it is isomorphic to the integral realoctonions of norm 1
modulo the center (cf. [5], [14]).
Let us recall loops of type M (G, 2) = M(G), first constructed in [1]. For a
2n
group G of order n, define new multiplication · on G×C by
2
(g, i)·(h, j)=((g(−1)jh(−1)i+j)(−1)j, i+j).
Then(G×C , ·)isaMoufangloop,andwedenoteitbyM(G). Itisnonassociative
2
if and only if G is nonabelian (cf. [1]).
Write M(G) = G∪Gu for some element u ∈ M(G)\G. Lemma 3.1 (cf. [16,
Prop 4.12]) is easy to prove once you realize that
- every element of Gu is of order 2,
- G·G=Gu·Gu=G, G·Gu=Gu·G=Gu,
- every subloop H 6≤G of M(G) satisfies |H ∩G|=|H ∩Gu|.
Lemma 3.1. Let G be a group of order n, and let M(G)=G∪Gu be constructed
as above.
(i) We have
H (C ), if m6=2,
H (C )= G m
M(G) m (cid:26) HG(Cm)+n, if m=2.
(ii) hH, gui∼=E2k+1 for every g ∈G, H ≤G, H ∼=E2k, k ≥0.
(iii) For k ≥1,
0, if 2k−1 ∤n,
H (E )=
M(G) 2k (cid:26) HG(E2k)+HG(E2k−1)·n·21−k, otherwise.
(iv) hg, hui∼=S for every g, h∈G with |g|=3.
3
(v) When H (C )6=0 and H (S )=0, then H (G)=1.
G 3 G 3 M(G)
It was proved in [15] that M(G) is presented (in the variety of Moufang loops)
by
hx, y, u; R, u2 =(xu)2 =(yu)2 =((xy)u)2 =ei, (8)
6 PETRVOJTEˇCHOVSKY´
whenever G is a 2-generatedgroup with presentation
hx, y; Ri.
In particular, presentations for the loops M(S ) and M(A ) can be obtained from
3 4
this result.
4. Main goal
We proceed to describe the subloop lattice of M∗(2), guided by steps (ℓ ), (ℓ ),
1 2
(ℓ ) of Section 1. We fulfil (ℓ ) and give reasonable amount of details with respect
3 1
to (ℓ ) and (ℓ ). In particular, we calculate all Hasse constants H (B), H (A|B),
2 3 C C
H∗(A|B), for A<B ≤C =M∗(2).
C
At several places, the reader will be kindly asked to verify a few details by
straightforward, easy calculations. Most of these calculations are reduced to a
quick glance into Table 1. The table itself can be checked for accuracy within
minutes, using Lemma 7.1. No machine computation is needed.
5. Possible subloops
Fix F = GF(2) and C = M∗(2). It is easy to see that C consists of 120 elements
of order 1, 2, 3. More precisely,
a α
x=
(cid:18) β b (cid:19)
satisfies |x|=2 if and only if a=b and x6=e; and |x|=3 if and only if a6=b. To
linearize our notation, we write x = [α, β] when |x| = 2, and x = {α, β} when
a a
|x| = 3. Since a can be calculated from α, β when |x| = 2, we further simplify
involutions to x = [α, β]. Note that {α, β}−1 = {α, β} , where the addition is
a 1+a
modulo 2.
Elementary counting reveals that there are 63 involutions and 56 elements of
order 3 in C. Using the language of Hasse constants, H (C ) = 63, H (C ) =
C 2 C 3
56/2=28.
CheinclassifiedallnonassociativeMoufangloopsoforderatmost63[2],andwe
willcallsuchMoufangloopssmall. SinceC has120elements,everypropersubloop
of C is small and can be found in Chein’s list.
As in [12], we say that a finite loop L has the weak Cauchy property when
it contains a subloop of order p for every prime p dividing |L|. It has the weak
Lagrange property if |H| divides |L| for every H ≤ L. Finally, L has the strong
Cauchy (Lagrange) property ifeverysubloopofLhastheweakCauchy(Lagrange)
property.
Since 5 divides |C|, C does not have the weak Cauchy property. However, it
followsfrom[2,Ch.XIV]thatallsmallMoufangloopshaveit. (Theyalsohavethe
strongLagrangeproperty. If one provesthat everyM∗(q) has the strong Lagrange
property,itwillfollowthatallMoufangloopshaveit(cf.[4]). Asofnow,this is an
SUBALGEBRAS AND HASSE CONSTANTS 7
open question. Glauberman proved [8] that all Moufang loops of odd order have
the strong Lagrange property.)
Corollary 5.1. The order of every proper subloop of C is 2r3s, for some r, s.
Lemma 5.2. Let x, y ∈ C, |x| = |y| = 3, y 6∈ hxi. Then hx, yi contains an
involution.
Proof. We may assume that x = {α, β} , y = {γ, δ} , for some α, β, γ, δ ∈ F3.
1 1
Then exactly one of the two elements xy, x2y is of order 2. (cid:3)
This means that 9 does not divide the order of any subgroupof C. Every group
oforder24containsanelement oforderatleast4 (the only twononabeliangroups
of order 24 with Sylow 2-subgroups isomorphic to E are D ×C and A ×C ).
8 6 2 4 2
Hence |G| ∈ {1, 2, 3, 4, 6, 8, 12, 16, 32, 48} for every subgroup G of C. It is
not obvious, at least to the author, that C contains no subgroups of order 16,
necessarily isomorphic to E . It is true, however, and we prove it in Section 14.
16
Hence |G|={1, 2, 3, 4, 6, 8, 12}.
Chein concludes in [2, Ch. XII] that every small Moufang loop containing no
elementofordergreaterthat3isnecessarilyoftheformM(G)forsomenonabelian
group G. Thanks to the restrictions on |G|, there are only two candidates for G,
namelyS (the symmetricgroupoforder6)andA (thealternatinggroupoforder
3 4
12).
Corollary 5.3. A nontrivial subloop of C is isomorphic to
C , C , E , S , E , A , M(S ) or M(A ). (9)
2 3 4 3 8 4 3 4
In particular, C has the strong Lagrange property.
All loops listed in (9) actually occur as subloops of C, as we shall see.
6. Automorphisms
We construct three kinds of automorphisms of C.
Lemma 6.1. Let f : F3 → F3 be a nonsingular linear transformation. Define
f :O(2)→O(2) by
a α a f(α)
b f = .
(cid:18) β b (cid:19) (cid:18) f(β) b (cid:19)
b
Then f ∈ Aut(O(2)) if (and only if) f is an automorphism of the Lie algebra
(F3, +, ×).
b
Proof. Linearity is obvious and the rest follows by straightforward computation
using Zorn’s multiplication (7). (cid:3)
Identify π ∈S with the linear transformationF3 →F3, (α , α , α )7→(α ,
3 1 2 3 π(1)
α ,α ). ByLemma6.1, π ∈Aut(C). Tokeepthe terminologysimple,wewill
π(2) π(3)
call such automorphisms permutations (of coordinates).
b
8 PETRVOJTEˇCHOVSKY´
Define ∂ :O(2)→O(2) by
a α b β
∂ = , (10)
(cid:18) β b (cid:19) (cid:18) α a (cid:19)
and verify that ∂ ∈Aut(O(2)).
Finally, we focus on conjugations. Not every conjugation in a Moufang loop is
an automorphism.
Lemma 6.2. Let Q be a simple Moufang loop. For x ∈ Q define γ : Q → Q by
x
γ (y)=x−1yx. Then γ is a nontrivial automorphism of Q if and only if |x|=3.
x x
Proof. By[12,ThmIV.1.6],γ isapseudo-automorphismwithcompanionx−3. So
x
γ is an automorphism whenever |x| divides 3.
x
Conversely, if γ is a nontrivial automorphism of Q, then it must be a pseudo-
x
automorphism with companions e and x−3. By [12, Thm IV.1.8], the set of all
companions of γ equals eN(Q), where N(Q) is the nucleus of Q. Since Q is
x
simple, we must have x−3 =e. (cid:3)
Remark 6.3. TheconclusionofLemma6.1remainsvalidoveranyfinitefieldGF(q),
butwethenget−π ∈Aut(O(q)), ratherthanπ ∈Aut(O(q)). Themap∂ :O(q)→
O(q) defined by (10) is an automorphism if and only if q is even.
c b
7. Subloops isomorphic to C
2
The detailed discussion concerning Sub(C) starts here.
Lemma 7.1. Let x = [α, β] , y = [γ, δ] be two involutions, x 6= y, and let
n m
z ={ε, ϕ} be an element of order 3 in C. Then:
t
(i) |xy|=2 if and only if hx, yi∼=E if and only if α·δ =β·γ,
4
(ii) |xy|=3 if and only if hx, yi∼=S if and only if α·δ 6=β·γ,
3
(iii) x is contained in a copy of S ,
3
(iv) every copy of S contains an involution of the form [ , ] ,
3 0
(v) |zx|=2 if and only if α·ϕ+β·ε=n,
(vi) z is contained in a copy of S .
3
Proof. The involution x commutes with y if and only if |xy|=2. Since
nm+α·δ
xy = ,
(cid:18) nm+β·γ (cid:19)
parts (i) and (ii) follow. Part (v) is proved similarly.
Let x = [α, β] . Without loss of generality, assume that β 6= 0. Pick γ, δ so
n
that α·δ = 0, β ·γ 6= 0. Then choose m ∈ {0, 1} so that y = [γ, δ] ∈ C. Then
m
hx, yi∼=S , and (iii) is proved.
3
Let G≤C, G∼=S , andsuppose that x=[α, β] , y =[γ, δ] ∈G, x6=y. Then
3 1 1
1+α·δ α+γ+β×δ
xy = .
(cid:18) β+δ+α×γ 1+β·γ (cid:19)
SUBALGEBRAS AND HASSE CONSTANTS 9
Since |xy| = 3, we have α·δ 6= β ·γ. In other words, α·δ+β ·γ = 1. Then the
third involution xyx∈G equals
1+α·δ+(α+γ)·β α·β
= .
(cid:18) (cid:19) (cid:18) (cid:19)
Now, α·β =0 since detx=1, and we are through with (iv).
Since detz 6=0,we canassumethatthe firstcoordinateofbothεandϕis equal
to1. Thenα=(1,0,0),β =(0,0,0)andn=1makex=[α, β] intoaninvolution
n
satisfying |zx|=2, by (v). This proves (vi). (cid:3)
Let us write α α α for the vector α = (α , α , α ) ∈ F3, and w(α) for α +
1 2 3 1 2 3 1
α +α (the weight of α).
2 3
Introduce x =[111, 111] as the canonical involution of C.
0
Observe that when x, y are two involutions of C generating a subgroup isomor-
phic to S then γ ∈Aut(C) maps x to y.
3 yx
Proposition 7.2. Aut(C) acts transitively on the 63 copies of C in C.
2
Proof. We show how to map an arbitraryx=[α, β] onto x . By Lemma 7.1(iii),
n 0
x is contained in a copy of S . Then, by Lemma 7.1(iv) and the observation
3
immediately preceding this Proposition, we can assume that n=0.
Let r = w(α), s = w(β). Using ∂ from Section 6, we can assume that r ≥ s.
We now fix y = [100, 100] and proceed to transform x into x′ so that x′ = x , or
0
x′ =y, or hx′, x i∼=S , or hx′, yi∼=S .
0 3 3
When r 6≡ s (mod 2), then hx, x i ∼= S , by Lemma 7.1(ii). So assume that
0 3
r ≡ s. Since n = 0, we have s > 0, and thus (r, s) = (1, 1), (2, 2), (3, 1), or
(3, 3). Everypermutation ofcoordinatescan be made into anautomorphismof C,
aswehaveseeninSection6. Moreover,x isinvariantunderallsuchpermutations.
0
When (r, s) = (1, 1), transform x into y. When (r, s) = (2, 2), transform x into
x′ =[110, 011], and note that hx′, yi∼=S . When (r, s)=(3, 1), transform x into
3
x′ =[111, 001], and note again that hx′, yi∼=S . Finally, when (r, s)=(3, 3), we
3
have x=x .
0
Now, when hx′, x i ∼= S or hx′, yi ∼= S , we can permute the involutions of C
0 3 3
so that x′ is mapped to x or y, respectively. Since x = γ (y), we are
0 0 {001,101}1
done. (cid:3)
Note that, inspiritof(ℓ ), the proofofProposition7.2tells us howto construct
3
automorphisms mapping involutions of C onto the representative x .
0
8. Subloops Isomorphic to C or S
3 3
Inthissection,weapplyProposition2.3forthefirsttime. Wehavetakenadvantage
of the fact that all permutations and ∂ leave x invariant. To proceed further, we
0
need additional automorphisms with this property.
Consider v = {010, 110} , v = {001, 101} , and define ξ : C → C by ξ =
0 0 1 0
γv1−1 ◦γv0. Then ξ ∈Aut(C), by Lemma 6.2, and ξ(x0)=x0.
10 PETRVOJTEˇCHOVSKY´
Set x = [110, 100], and let y = x x = {011, 110} be the canonical element
1 0 0 1 1
of order 3.
Proposition 8.1. Aut(C) acts transitively on the copies of S and C .
3 3
Proof. Since H (C ) = 1 and H (C |S ) > 0, by Lemma 7.1(vi), it suffices to
S3 3 C 3 3
prove that Aut(C) acts transitively on the copies of S . Let G ∼= S , G = hx, yi,
3 3
|x| = |y| = 2. By Proposition 7.2, we can assume that x = x . Write y = [α, β] ,
0 n
r = w(α), s =w(β). By Lemma 7.1, we have r 6≡ s. Using ∂, we can assume that
r>s. We are going to transform y into x .
1
Assumethatn=1. Thenα·β =0. Takingpermutationsofcoordinatesandthe
possible values of (r, s) into account, we transform y into one of x = [010, 000],
2
x = [011, 100], x = [111, 000], x = [111, 101]. With ξ as above, check that all
3 4 5
of ξ(x ), ξ−1(x ), ξ(x ) and ξ−1(x ) have zeros on the diagonal.
2 3 4 5
We may hence assume that n = 0. Then (r, s) = (2, 1), and we are done by
permuting coordinates. (cid:3)
Lemma 8.2. H (C |S )=16, H (S )=336, H (C |S )=12.
C 2 3 C 3 C 3 3
Proof. Pickaninvolutionx. ByProposition8.1,thenumberofinvolutionsysatisfy-
ing|xy|=3isindependentofx. Onecanthenimmediatelyseewithx=[100, 100],
say, that there are 32 such involutions. As H (C ) = 3, we get H (C |S ) = 16.
S3 2 C 2 3
Then, by (5), H (S ) = H (C )·H (C |S )·H (C )−1 = 336. Again by (5),
C 3 C 2 C 2 3 S3 2
H (C |S )=H (C )·H (S )·H (C )−1 =12. (cid:3)
C 3 3 S3 3 C 3 C 3
Note that Lemma 7.1 allows us to construct all copies of S containing x , and
3 0
also all copies of S containing y . Note further that we did not have to resort to
3 0
local analysis to find the value of H (C |S ).
C 3 3
From this moment on, we will pay less attention to (ℓ ) and (ℓ ).
2 3
9. Subloops isomorphic to A
4
Fix z ={110, 100} , and recall that H (C )=3, H (C )=4.
0 0 A4 2 A4 3
Proposition 9.1. Aut(C) acts transitively on the 63 copies of A , and
4
H (C |A )=3.
C 2 4
Proof. Working in C, we have hG, xi∼=S or A for every copy G of C and every
3 4 3
involution x. Since H (C |S ) = 12 and H (C ) = 3, there are 36 involutions x
C 3 3 S3 2
in G such that hG, xi ∼= S3. Thus HC(C3|A4) = (63−36)·HA4(C2)−1 = 9. By
(5), H (A )=H (C )·H (C |A )·H (C )−1 =63.
C 4 C 3 C 3 4 A4 3
As for the transitivity, pick G ∼= A . By Proposition 7.2, we can assume that
4
G = hx , zi, for some z = {ε, ϕ} with r = w(ε), s = w(ϕ). Since |x z| = 3,
0 t 0
we have r 6≡ s, by Lemma 7.1(v), and may thus assume that r > s ≥ 1. Then
(r, s)=(2, 1) is the only possibility, and z can be transformed to z or z−1. (cid:3)
0 0
Perhaps it would be more natural to look at the copies of E now, however,
4
the Klein subgroups of C are exceptional in the sense that Aut(C) does not act